携程后台笔试，求大佬指教指教

// 给定一个单向链表和一个整数m，将链表中小于等于m的节点移到大于m的节点之前，要求两部分中的节点各自保持原有的先后顺序
// 样例输入
// 4
// 9 6 3 7 6 5
// 样例输出
// 3,9,6,7,6,5

static ListNode partition(ListNode head,int m) {
        if(head==null)
            return null;
        ListNode temp=new ListNode(m),newHead=null,p1=new ListNode(-1),p2=new ListNode(-1);
        p1.next=temp;
        p2=temp;
        while(head!=null){
            ListNode next=head.next;
            if(head.val<=m){
                p1.next=head;
                head.next=temp;
                p1=head;
                if(newHead==null)
                    newHead=head;
            }else{
                p2.next=head;
                p2=p2.next;
            }
            head=next;
        }
        p1.next=temp.next;
        return newHead;
    }

//	豚厂给自研的数据库设计了一套查询表达式，在这个表达式中括号表示将里面的字符串翻转。请你帮助实现这一逻辑。括号不匹配，输出空字符	
// 样例输入
//	((ur)oi)
//	样例输出
//	iour
	static String resolve(String expr) {
        if(expr==null)
            return null;
        Stack<Integer> stack1=new Stack<>();
        Stack<Integer> stack2=new Stack<>();
        for(int i=0;i<expr.length();i++){
            char c=expr.charAt(i);
            if(c!=')'){
                stack1.add((int)c);
            }else{
                if(!stack1.isEmpty()){
                    while(stack1.peek()!=(int)'(')
                        stack2.add(stack1.pop());
                    stack1.pop();
                    while(!stack2.isEmpty()){
                        stack1.add(stack2.pop());
                    }
                }
            }
        }
        String rs="";
        boolean match=true;
        while(!stack1.isEmpty()){
            if(stack1.peek()=='('||stack1.peek()==')'){
                match=false;
                break;
            }
            rs+=stack1.pop();
        }
        return match?rs:"";
    }